x Turn on thread page Beta
 You are Here: Home >< Maths

# Show that arctan(x) exists and is continuous watch

1. Show tan(x): (-pi/2, pi/2) -> R has a continuous inverse arctan(x) : R -> (-pi/2, pi/2).

You may assume that tan(x) is continuous and strictly increasing on the given domain, and tends to +/- at +/- pi/2

I think I have shown that tan(x) has an inverse on this domain by showing it is bijective.

However, I am unsure how to go about showing that arctan(x) is continuous.
2. What have you tried so far?
3. I assumed not, negating the definition of continuous, and chose epsilon(e)=4, then said the maximum of f(p)-f(x) was pi (pi/2 - (-pi/2)), so there is no such |f(x) - f(p)|>=4. However, I do not think this is valid as epsilon is bigger than the domain of arctan.
4. What do you think the negation of the definition of continuous is? (Because from what you have written, you don't seem to have it right).
5. There exists p in R, there exists e>0 for all d>0 there exists x in R such that |x-p|<d implies |f(x)-f(p)| >=e
6. Mostly right, but "there exists x in R such that |x-p|<d implies |f(x)-f(p)| >=e" should be " with |x-p| < d and ".

Anyhow, that's not your first problem. You are doing a proof by contradiction, so you need to show that your statement (that f is not continuous) is *never* true. In other words, you need to show it doesn't work *whatever* you choose for epsilon. You don't get to choose it yourself.
7. Ah yes, that is true. I have to hand it in now anyway so I'll just wait and see how it is explained but thanks for spotting my errors
8. (Original post by DFranklin)
Mostly right, but "there exists x in R such that |x-p|<d implies |f(x)-f(p)| >=e" should be " with |x-p| < d and ".
Im interested for you to explain the difference between these two definitions
9. (Original post by Tompazr)
Im interested for you to explain the difference between these two definitions
For the first one; it's basically saying there exists x, such that A implies B. You could choose an x that makes A false, and it will satisfy statement; since if A is false, then A implies B is true.

Which is completely different from there exists x such that A and B, which can only be satisfied if A is satisfied, and B is satisfied.

Not terribly well worded, but hope that made sense.
10. Could you say that since is continuous and has a well defined inverse, both of which forms part of a group then all operations on the group maintain continuity then hence any combination of e^x is still continuous? Or is that just rubbish?
11. Um, rubbish, I'm afraid.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: February 9, 2010
Today on TSR

### How do I turn down a guy in a club?

What should I do?

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams